Đáp án:
Giải thích các bước giải:
a)
Xét $\triangle AHF$ và $\triangle ACH$:
$\hat{AFH}=\hat{AHC}(=90^o)$
$\hat{HAC}$: chung
$\to \triangle AHF\backsim \triangle ACH$ (g.g)
$\to \dfrac{AF}{AH}=\dfrac{AH}{AC}\to AF.AC=AH^2$
Xét $\triangle AEH$ và $\triangle AHB$:
$\hat{AEH}=\hat{AHB}(=90^o)$
$\hat{HAB}$: chung
$\to \triangle AEH\backsim \triangle AHB$ (g.g)
$\to \dfrac{AE}{AH}=\dfrac{AH}{AB}\to AE.AB=AH^2$
$\to AE.AB=AF.AC(=AH^2)$
b)
Ta có: $AE.AB=AF.AC \to \dfrac{AE}{AC}=\dfrac{AF}{AB}$
Xét $\triangle AEF$ và $\triangle ACB$:
$\hat{EAF}=\hat{CAB}(=90^o)$
$\dfrac{AE}{AC}=\dfrac{AF}{AB}$ (cmt)
$\to \triangle AEF\backsim\triangle ACB$ (c.g.c)
c)
Xét $\triangle ABH$ và $\triangle CBA$:
$\hat{AHB}=\hat{CAB}(=90^o)$
$\hat{CBA}$: chung
$\to \triangle ABH\backsim\triangle CBA$ (g.g)
$\to \dfrac{AB}{CB}=\dfrac{BH}{BA}$
$\to CB.BH=AB^2 \to (CH+HB).BH=AB^2 \to (4+HB)BH=(\sqrt{12})^2$
$\to BH^2+4BH-12=0 \to (BH-2)(BH+6)=0 \to \left[\begin{array}{l}BH=2 \text{ (thoả mãn)}\\BH=-6\text{ (loại)}\end{array}\right.\to BH=2cm$
$\to BC=CH+HB=4+2=6cm$
Ta có: $AB^2+AC^2=BC^2\to AC=\sqrt{BC^2-AB^2}=\sqrt{6^2-(\sqrt{12})^2}=2\sqrt{6}cm$
